Difference between revisions of "2014 AMC 10B Problems/Problem 9"
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+ | ==Solution 4== | ||
+ | |||
+ | Set | ||
+ | |||
+ | Substitute the new values into the first equation | ||
+ | |||
+ | <math>1/2 + 1 = 3/2</math>, | ||
+ | |||
+ | <math>1/2 - 1 = -1/2</math>, | ||
+ | |||
+ | <math>(3/2) / (-1/2) = -3</math> | ||
+ | |||
+ | Substitute in the second equation with new values of | ||
+ | |||
+ | (2 + 1) / (2 - 1) = 3. | ||
+ | |||
+ | Answers of each equation (where X is the quotient): <math>x</math> and <math>-x</math> | ||
+ | |||
+ | Therefore, the answers to the equations are the negatives of each other. Thus, | ||
+ | |||
+ | ~WalkEmDownTrey | ||
==Video Solution (CREATIVE THINKING)== | ==Video Solution (CREATIVE THINKING)== |
Revision as of 21:47, 5 October 2024
Contents
[hide]Problem
For real numbers and , What is ?
Solution
Multiply the numerator and denominator of the LHS (left hand side) by to get . Then since and , , or choice .
Solution 2
Muliply both sides by to get . Then, add to both sides and subtract from both sides to get . Then, we can plug in the most simple values for z and w ( and , respectively), and find , or answer choice .
Solution 3
Let and . To find values for a and b, we can try and . However, that leaves us with a fractional solution, so scaling it by 2, we get and . Solving by adding the equations together, we get and . Now, substituting back in, we get and . Now, putting this into the desired equation with (since it will cancel out), we get . Dividing, we get .
~idk12345678
Solution 4
Set
Substitute the new values into the first equation
,
,
Substitute in the second equation with new values of
(2 + 1) / (2 - 1) = 3.
Answers of each equation (where X is the quotient): and
Therefore, the answers to the equations are the negatives of each other. Thus,
~WalkEmDownTrey
Video Solution (CREATIVE THINKING)
~Education, the Study of Everything
Video Solution
~savannahsolver
See Also
2014 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 8 |
Followed by Problem 10 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AMC 10 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.